5+10+...+20.6+2020
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\(A=\frac{2^{12}\cdot27^3+20\cdot6^9}{2\cdot6^{10}+12^6\cdot3^5}\)
\(=\frac{2^{12}\cdot\left(3^3\right)^3+2^2\cdot5\cdot2^9\cdot3^9}{2\cdot2^{10}\cdot3^{10}+\left(2^2\right)^6\cdot3^6\cdot3^5}\)
\(=\frac{2^{12}\cdot3^9+2^{11}\cdot5\cdot3^9}{2^{11}\cdot3^{10}+2^{12}\cdot3^{11}}\)
\(=\frac{2^{11}\cdot3^9\left(2\cdot5\right)}{2^{11}\cdot3^{10}\left(2\cdot3\right)}\)
\(=\frac{2^{11}\cdot3^9\cdot10}{2^{11}\cdot3^{10}\cdot6}\)
\(=3\cdot\frac{10}{6}=\frac{30}{6}=5\)
a)
\(\begin{array}{l}2023 - {25^2}:{5^3} + 27\\ = 2023 - {5^4}:{5^3} + 27\\ = 2023 - 5^{4-3} + 27\\ = 2023 -5 + 27\\ =2018+27\\= 2045\end{array}\)
b)
\(\begin{array}{l}60:\left[ {7.\left( {{{11}^2} - 20.6} \right) + 5} \right]\\ = 60:\left[ {7.\left( {121 - 120} \right) + 5} \right]\\ = 60:\left[ {7.1 + 5} \right]\\ = 60:12\\ = 5\end{array}\)
A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)
10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2
B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)
10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2
10A > 2 > 10B ⇒ 10A>10B ⇒ A>B
ta có :
A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)
B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)
\(\Leftrightarrow\) B < A